Deriving the Gamma Function

Date: 12/15/2000 at 01:53:41
From: James Garcioparra
Subject: Calculus/Probability/Statistics

How can you prove that the square root of pi divided by 2 is 0.5 
factorial? What is a fractional factorial like that equal to?

Date: 12/15/2000 at 19:02:14
From: Doctor Anthony
Subject: Re: Calculus/Probability/Statistics

Factorials can be generated by the Gamma function, and it is the Gamma 
function that allows us to define things like (1/2)!

The Gamma function is defined as:

             0
     G(n) = INT [x^(n-1) e^(-x) dx]
            inf

Now integrating by parts

   G(n) = [x^(n-1).(-e^(-x)] + INT[(n-1)x^(n-2).e^(-x).dx]

The first bracket is zero at both ends when limits infinity and 0 are 
put in. So:

     G(n) = (n-1)INT[x^(n-2).e^(-x).dx]

     G(n) = (n-1).G(n-1)  ........................................[1]

and when n is a positive integer this defines the factorial function.

     G(n) = (n-1)(n-2)(n-3) ... 3 * 2 * 1 * G(1)

and

             0
     G(1) = INT [e^(-x)]
            inf

                     | 0
         = -[e^(-x)] |
                     | inf

         = -[0 - 1]

         = 1

It follows that:

     G(n) = (n-1)!

It follows also that:

     G(1) = (1-1)!

          = 0!

and we have seen that G(1) = 1, so then 0! = 1 .

Now looking at non-integer values for the gamma function, we have for 
example:

                0
      G(1/2) = INT [x^(-1/2).e^(-x).dx]
               inf

Putting x = u^2 and integrating this gives:

                 0
     G(1/2) = 2.INT [e^(-u^2).du]
                inf

This is the well-known integral for the normal distribution and we 
have:

     G(1/2) = sqrt(pi)

Using this result we can now obtain all other positive half-integral 
values from the recurrence relation [1]. For example:

     G(3/2) = (1/2).G(1/2) = sqrt(pi)/2

     G(7/2) = (5/2).G(5/2) = (5/2)(3/2).G(3/2) = (15/4).sqrt(pi)/2

The recurrence relation [1] can also be used in defining the gamma 
function for negative values of n. Rewriting [1] as G(n) = G(n+1)/n, 
we have:

     G(-3/2) = G(-1/2)/(-3/2)

             = G(1/2)/[(-3/2)(-1/2)]

             = (4/3).sqrt(pi)

For other values such as G(1/3) or whatever we have tables of values 
derived from integrating series approximations to the function.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/